Question: Solve for $z$, $ -\dfrac{6}{z - 3} = -\dfrac{z - 6}{z - 3} - \dfrac{10}{2z - 6} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $z - 3$ $z - 3$ and $2z - 6$ The common denominator is $2z - 6$ To get $2z - 6$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{z - 3} \times \dfrac{2}{2} = -\dfrac{12}{2z - 6} $ To get $2z - 6$ in the denominator of the second term, multiply it by $\frac{2}{2}$ $ -\dfrac{z - 6}{z - 3} \times \dfrac{2}{2} = -\dfrac{2z - 12}{2z - 6} $ The denominator of the third term is already $2z - 6$ , so we don't need to change it. This give us: $ -\dfrac{12}{2z - 6} = -\dfrac{2z - 12}{2z - 6} - \dfrac{10}{2z - 6} $ If we multiply both sides of the equation by $2z - 6$ , we get: $ -12 = -2z + 12 - 10$ $ -12 = -2z + 2$ $ -14 = -2z $ $ z = 7$